Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

A chemical is diluted out of a tank by pumping pure water into the tank and pumping the existing solution out of it, so the volume at any time
t is
20+2t.
The amount
z of chemical in the tank decreases at a rate proportional to
z and inversely proportional to the volume of solution in the tank.
Which equation describes this relationship?
Choose 1 answer:
(A)
(dz)/(dt)=kz-(1)/(20+2t)
(B)
(dz)/(dt)=k(20+2t)-(1)/(z)
(c)
(dz)/(dt)=-(k(20+2t))/(z)
(D)
(dz)/(dt)=-(kz)/(20+2t)

A chemical is diluted out of a tank by pumping pure water into the tank and pumping the existing solution out of it, so the volume at any time $t$ is $20+2 t$ . $\newline$ The amount $z$ of chemical in the tank decreases at a rate proportional to $z$ and inversely proportional to the volume of solution in the tank. $\newline$ Which equation describes this relationship? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{d z}{d t}=k z-\frac{1}{20+2 t}$ $\newline$ (B) $\frac{d z}{d t}=k(20+2 t)-\frac{1}{z}$ $\newline$ (c) $\frac{d z}{d t}=-\frac{k(20+2 t)}{z}$ $\newline$ (D) $\frac{d z}{d t}=-\frac{k z}{20+2 t}$

View step-by-step help

View step-by-step help

One-step inequalities: word problems

Full solution

Q. A chemical is diluted out of a tank by pumping pure water into the tank and pumping the existing solution out of it, so the volume at any time

t

is

20+2 t

.

\newline

The amount

z

of chemical in the tank decreases at a rate proportional to

z

and inversely proportional to the volume of solution in the tank.

\newline

Which equation describes this relationship?

\newline

Choose

1

answer:

\newline

(A)

\frac{d z}{d t}=k z-\frac{1}{20+2 t}

\newline

(B)

\frac{d z}{d t}=k(20+2 t)-\frac{1}{z}

\newline

(c)

\frac{d z}{d t}=-\frac{k(20+2 t)}{z}

\newline

(D)

\frac{d z}{d t}=-\frac{k z}{20+2 t}

Identify Proportional Relationship: The problem states that the rate of decrease of the chemical $z$ is proportional to $z$ and inversely proportional to the volume of the solution in the tank, which is $20 + 2t$ . This means that the rate of change of $z$ with respect to time $t$ , denoted as $\frac{dz}{dt}$ , should be equal to a constant $k$ times $z$ , and divided by the volume $20 + 2t$ . This gives us the equation $\frac{dz}{dt} = -\frac{kz}{20+2t}$ , where the negative sign indicates that $z$ is decreasing.
Derive Rate Equation: We can now compare the derived equation with the given options to find the correct answer. $\newline$ Option (A) $\frac{dz}{dt} = kz - \frac{1}{20+2t}$ does not match our derived equation because it suggests that the rate of decrease is not inversely proportional to the volume.
Compare with Options: Option (B) $\frac{dz}{dt} = k(20+2t) - \frac{1}{z}$ also does not match our derived equation because it suggests that the rate of decrease is directly proportional to the volume, which is not the case.
Match with Correct Option: Option (C) $\frac{dz}{dt} = -\frac{k(20+2t)}{z}$ does not match our derived equation because it suggests that the rate of decrease is inversely proportional to $z$ , which is not the case.
Match with Correct Option: Option (C) $\frac{dz}{dt} = -\frac{k(20+2t)}{z}$ does not match our derived equation because it suggests that the rate of decrease is inversely proportional to $z$ , which is not the case.Option (D) $\frac{dz}{dt} = -\frac{kz}{20+2t}$ matches our derived equation perfectly, indicating that the rate of decrease of the chemical is proportional to the amount of chemical $z$ and inversely proportional to the volume $20 + 2t$ .

More problems from One-step inequalities: word problems

Question

-11 x \geq-33

\newline

Which of the following best describes the solutions to the inequality shown?

\newline

1

\newline

x \leq-3

\newline

x \geq-3

\newline

x \leq 3

\newline

x \geq 3

Posted 1 year ago

Question

\frac{3}{2} q \leq \frac{9}{2} q-18

\newline

Which of the following best describes the solutions to the inequality shown?

\newline

1

\newline

q \leq 3

\newline

q \geq 3

\newline

q \leq 6

\newline

q \geq 6

Posted 1 year ago

Question

g(x)=5-e^{x}

\newline

Below is Sean's attempt to write a formal justification for the fact that the equation

g(x)=0

has a solution where

1 \leq x \leq 4

\newline

Is Sean's justification complete? If not, why?

\newline

Sean's justification:

\newline

g

is defined for all real numbers, and exponential functions are continuous at all points in their domains.

\newline

So, according to the intermediate value theorem,

g(x)=0

must have a solution somewhere between

x=1

x=4

\newline

1

\newline

(A) Yes, Sean's justification is complete.

\newline

(B) No, Sean didn't establish that

0

g(1)

g(4)

\newline

(C) No, Sean didn't establish that

g

Posted 1 year ago

Question

f(x)=\sqrt[3]{\cos (\pi \cdot x)}

\newline

Below is Issac's attempt to write a formal justification for the fact that the equation

f(x)=0.1

has a solution where

1.5 \leq x \leq 2

\newline

Is Issac's justification complete? If not, why?

\newline

Issac's justification:

\newline

\begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1<f(2) . \end{array}

\newline

So, according to the intermediate value theorem,

f(x)=0.1

must have a solution when

x

x=1.5

x=2

\newline

1

\newline

(A) Yes, Issac's justification is complete.

\newline

(B) No, Issac didn't establish that

0

1

f(1.5)

f(2)

\newline

(C) No, Issac didn't establish that

f

Posted 1 year ago

Question

h(x)=5 \cdot \tan (x)

\newline

Below is Bridget's attempt to write a formal justification for the fact that the equation

h(x)=2

has a solution where

0 \leq x \leq \frac{\pi}{4}

\newline

Is Bridget's justification complete? If not, why?

\newline

Bridget's justification:

\newline

h

is defined over the entire interval

\left[0, \frac{\pi}{4}\right]

, and trigonometric functions are continuous at all points in their domains.

\newline

So, according to the intermediate value theorem,

h(x)=2

must have a solution at some point in that interval.

\newline

1

\newline

(A) Yes, Bridget's justification is complete.

\newline

(B) No, Bridget didn't establish that

2

h(0)

h\left(\frac{\pi}{4}\right)

\newline

(C) No, Bridget didn't establish that

h

Posted 1 year ago

Question

f(x)=2^{5-2 x}

\newline

Below is Ibrahim's attempt to write a formal justification for the fact that the equation

f(x)=10

has a solution where

-1 \leq x \leq 4

\newline

Is Ibrahim's justification complete? If not, why?

\newline

Ibrahim's justification:

\newline

f

is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also,

f(-1)=128

f(4)=\frac{1}{8}

10

f(-1)

f(4)

\newline

So, according to the intermediate value theorem,

f(x)=10

must have a solution at some point between

x=-1

x=4

\newline

1

\newline

(A) Yes, Ibrahim's justification is complete.

\newline

(B) No, Ibrahim didn't establish that

10

f(-1)

f(4)

\newline

(c) No, Ibrahim didn't establish that

f

Posted 1 year ago

Question

g(x)=\cos \left(\pi x^{2}\right)

\newline

Below is Amrita's attempt to write a formal justification for the fact that the equation

g^{\prime}(x)=0.4

has a solution where

-1<x<4

\newline

Is Amrita's justification complete?

\newline

\newline

Amrita's justification:

\newline

Polynomial and trigonometric functions are differentiable and continuous at all points in their domain, and

[-1,4]

g^{\prime}

s domain. Also,

\newline

\begin{array}{l} g(-1)=-1 \text { and } \\ g(4)=1 \text {, so } \\ \frac{g(4)-g(-1)}{4-(-1)}=0.4 . \end{array}

\newline

So, according to the mean value theorem,

g^{\prime}(x)=0.4

must have a solution somewhere between

x=-1

x=4

\newline

1

\newline

(A) Yes, Amrita's justification is complete.

\newline

(B) No, Amrita didn't establish that the average rate of change of

g

[-1,4]

0

4

\newline

(C) No, Amrita didn't establish that

g

is differentiable.

Posted 1 year ago

Question

\newline

h(x)=x^{5}-2 x^{4}+10 x^{2}-10 x \text {. }

\newline

Below is Omar's attempt to write a formal justification for the fact that there exists a value

c

\newline

-2<c<2 \text { such that } h^{\prime}(c)=6 \text {. }

\newline

Is Omar's justification complete? If not, why?

\newline

Omar's justification:

\newline

\begin{array}{l} h(-2)=-4 \text { and } \\ h(2)=20 \text {, so } \\ \frac{h(2)-h(-2)}{2-(-2)}=6 . \end{array}

\newline

So, according to the mean value theorem, there exists a value

c

somewhere in the interval

-2<c<2

h^{\prime}(c)=6

\newline

1

\newline

(A) Yes, Omar's justification is complete.

\newline

(B) No, Omar didn't establish that the average rate of change of

h

[-2,2]

6

\newline

(C) No, Omar didn't establish that

h

is differentiable.

Posted 1 year ago

Question

f(x)=\sqrt{x+\sqrt{x^{3}}}

\newline

Below is Lena's attempt to write a formal justification for the fact that the equation

f^{\prime}(x)=\frac{2}{3}

has a solution where

0<x<9

\newline

Is Lena's justification complete? If not, why?

\newline

Lena's justification:

\newline

f(0)=0

f(9)=6

, so the average rate of change of

f

over the interval

0 \leq x \leq 9

\frac{2}{3}

\newline

So, according to the mean value theorem,

f^{\prime}(x)=\frac{2}{3}

must have a solution somewhere in the interval

0<x<9

\newline

1

\newline

(A) Yes, Lena's justification is complete.

\newline

(B) No, Lena didn't establish that the average rate of change of

f

[0,9]

\frac{2}{3}

\newline

(C) No, Lena didn't establish that

f

is differentiable.

Posted 1 year ago

Question

h(x)=3^{x}-x^{2}

\newline

Below is Uriah's attempt to write a formal justification for the fact that the equation

h^{\prime}(x)=8

has a solution where

1<x<3

\newline

Is Uriah's justification complete? If not, why?

\newline

Uriah's justification:

\newline

Polynomial and exponential functions are differentiable and continuous at all points in their domain, and

[1,3]

h

's domain. Also,

h(1)=2

h(3)=18

\newline

\frac{h(3)-h(1)}{3-1}=8 \text {. }

\newline

So, according to the mean value theorem,

h^{\prime}(x)=8

must have a solution somewhere between

x=1

x=3

\newline

1

\newline

(A) Yes, Uriah's justification is complete.

\newline

(B) No, Uriah didn't establish that the average rate of change of

h

[1,3]

8

\newline

(C) No, Uriah didn't establish that

h

is differentiable.

Posted 1 year ago

Related topics

Algebra - Order of Operations Algebra - Distributive Property `X` and `Y` Axes Geometry - Scalene Triangle Common Multiple Geometry - Quadrant